# Find a point on t

Given:

The curve y = x3 – 2x2 – 2x and a line y = 2x – 3

First, we will find The Slope of tangent

y = x3 – 2x2 – 2x

(x3) – (2x2) – (2x)

= 3x3 – 1 (x2 – 1) – 2x1 – 1

= 3x2 – 4x – 2 ...(1)

y = 2x – 3 is the form of equation of a straight line y = mx + c,where m is the The Slope of the line.

so the The Slope of the line is y = 2(x) – 3

Thus, The Slope = 2. ...(2)

From (1) & (2)

3x2 – 4x – 2 = 2

3x2 – 4x = 4

3x2 – 4x – 4 = 0

We will use factorization method to solve the above Quadratic equation.

3x2 – 6x + 2x – 4 = 0

3x(x – 2) + 2(x – 2) = 0

(x – 2)(3x + 2) = 0

(x – 2) = 0 & (3x + 2) = 0

x = 2 or

x =

Substitute x = 2 & x = in y = x3 – 2x2 – 2x

when x = 2

y = (2)3 – 2 (2)2 – 2(2)

y = 8 – (24) – 4

y = 8 – 8 – 4

y = – 4

when x =

y = ()3 – 2 ()2 – 2()

y = () – 2 () + ()

y = () – () + ()

taking lcm

y =

y =

y =

Thus, the points are (2, – 4) & (,)

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